• Complex Analysis: #28 Stirling`s Formula

The functional equation for the gamma function, Γ(z + 1) = zΓ(z), together with the observation that Γ(1) = 1, shows that for all n ∈ ℕ, we have Γ(n + 1) = n!. In the analysis lecture, I gave a very simple approximation to the gamma function. Namely, for large integers n, we have n! ≈ nⁿe⁻ⁿ. But this is really a very inexact approximation, which we were able to derive with hardly any thought at all. In this section, we will be looking at the “genuine” Stirling formula. It is

where P₁(t) = t − [t] − 1/2, and [t] is the largest integer which is less than or equal to t. Of course we take the principle value of the logarithm here. The fact that, as long as z is not a negative real number, the integral on the right-hand side of the equation goes to zero as |z| → ∞, shows that

(at least if we stay away from the negative real numbers). In fact we will show that for a given fixed δ between 0 and π, we have that P₁(z) falls uniformly to 0, for z = re^iθ and |θ| ≤ π − δ.

Lemma 1 (Euler’s Summation formula)
Let f : ℝ → ℝ be continuously differentiable (in the sense o real analysis). Then

Taking the sum from k = 1 to n then gives the result.

Similarly we have

Lemma 2
Assume z is not a negative real number or zero. Let P₂(t) = t_∗(t_∗ − 1)/2, where t_∗ = t − [t] for t ∈ ℝ. Then we have

The function Φ which is so defined is analytic in ℂ (minus the negative reals and 0).

Proof
Using partial integration, we have

where K is a finite number. (When z is a positive real number, the integral obviously also convergesThus we can use Weierstrass’ convergence theorem — on the integrals between successive integer— to conclude that Φ is an analytic function.

Lemma 3
We have

Proof
This follows from the previous lemma, since then K = 0 and π/2y → 0, when y → ∞.

So now that we have looked at these lemmas, let us begin proving Stirling’s formula. It involves the standard, rather awkward problem of the logarithms. So to avoid this, let us first take the case that z is just a positive real number. That is, we take z = x + i · 0, and we set about examining log Γ (x). Obviously Γ (x) > 0, so we can use the simple, real logarithm function. The antiderivative of log(x) is x log(x) − x. Therefore, using lemma 1, we see that for n ∈ ℕ we have

is simply some constant real number, let’s call it C. But then we see that the left-hand side of the equation, namely log Γ (x) is just the restriction to the positive real numbers of an analytic function which is defined in the region consisting of C with the negative real numbers and 0 removed. Similarly the right-hand side, namely xlog x − (log x)/2 − x − Φ(x) + C is the restriction to the positive real numbers of another analytic function which is defined in the same region. Since both of these functions coincide on the positive real numbers, they must be same analytic function.

So the only remaining problem is to find out what the value of C is. Writing

Γ(1 − z) = −zΓ(−z), 

we have

This is true for all y > 0, no matter how large. But according to lemma 3, we have lim_y→∞Φ(iy) = 0. For the rest, we should remember that for a complex number of the form re^iθ, we have

Complex Analysis: #27 Gamma Function II

• Complex Analysis: #27 The Gamma Function II

But what is the relationship with the formula

for x a real number greater than 1, which we used in the analysis lectures? Going from real to complex numbers, let us write z = x + iy. Then we have

Therefore — as we saw in Analysis I — the integral defining the gamma function will again converge when x > 1; for x < 1 it diverges.

As an exercise, we see that for z = x + iy and x > 0, we have

This is another standard formula for the gamma function, and in fact it is valid as long as Re(z) > 0. But we still haven’t shown that it is the same gamma function as that which was defined in the previous section.

For each n ∈ ℕ, let

and we see the connection with the definition in the previous section.

As a final remark before proceeding with Stirling’s formula, I should mention that Legendre introduced the beta function, which is

This seems to be used these days in computer algebra programs. Also, when on that subject, one finds many interesting and obscure formulas related to the gamma function. (You can also look in books containing tables of mathematical formulas.) One such formula which caught my eye is the following. Let n ∈ ℕ. Then

Here n!! is the double factorial. That is

Complex Analysis: #26 Gamma Function I

• Complex Analysis: #26 The Gamma Function I

In the analysis lecture, we defined the gamma function for real numbers x > 1 using the improper integral

We can do the same thing here in the realms of complex analysis. Just substitute the complex number z for the real number x in the formula. Using the same argument as in the real case, we find that the integral converges if Re(z) > 1. But by doing this, we miss out on the fact that the gamma function can be defined everywhere in  ℂ (with isolated poles). Of course it is again possible to look at the functional equation for the gamma function, then bring in analytic continuation. But this is a rather complicated manipulation! In reality, an infinite product representation is simpler.

Let’s begin by looking at the following function

We will be using the second expression here, although the first expression shows that we can apply theorem 48 and conclude that the product does converge to an entire function, whose zeros are simply the set of negative real integers. Taking a look at the product representation of the sine function which we obtained in the last section, we see that

The function zG(z) obviously has as it’s zeros the negative real integers, and also 0. But then, if we take the function G(z − 1), we see that it also has the negative real integers and 0 as it’s set of zeros. Since all of these zeros are simple zeros for both functions, we have

being an entire function which is never zero. It is now an exercise to show that there must exist an entire function γ : ℂ → ℂ such that

ze^γ(z)G(z) = G(z − 1) 

for all z ∈ ℂ.

Let’s take the logarithmic derivative on both sides of the equation. We get

so the sum is certainly absolutely convergent and thus the third equation is valid. But this means that γ '(z) = 0, that is, γ(z) is a constant, which we simply denote by γ. In fact it is Euler’s constant.

To see this, consider the case z = 1. From the formula defining G, we certainly have G(0) = 1. Therefore G(0) = 1 = e^γG(1), or

So G is an entire function satisfying the functional equation G(z−1) = ze^γG(z). In order to improve the appearance of things, let us define H(z) = G(z)e^z·γ. Then we have the functional equation

H(z − 1) = zH(z). 

This is beginning to look like the functional equation for the gamma function, but unfortunately (or fortunately?) it is going in the “wrong” direction. To fix this up, we take

From the construction, we see that it has simple poles at the negative integers, and at 0, but it has no zeros. Also, looking at the equation for the sine function, we can express this in terms of the gamma function:

• Complex Analysis: #25 Some Infinite Products

Genus zero:
Given a sequence of non-zero points a_n with ∑ 1/|a_n| converging, we can simply say that

gives us an example of a genus zero function with zeros just where we want them. Of course if we also want a zero at 0, then we can multiply this product with the factor Cz^l say, where l is the order of this zero, and C is a non-zero constant which we can choose as we like.


The sine function:
Being more concrete, let us look at the function sin πz. According to the definition of the sine function, we have

But this can only be zero if e^iπz  = e^−iπz. Writing z = x + iy, this means that e^−ye^iπx = e^ye^−iπx. In particular, y = 0 so that the zeros are just the familiar zeros which we know from real analysis, namely the integers, ℤ. As we know, the series ∑ 1/n² converges, thus the complex sine function must have genus one. Writing

the problem is then to determine the function g. For this, we take the logarithmic derivatives of both sides. We obtain

so that the sum is uniformly convergent in compact subsets of ℂ which do not contain points of ℤ, thus showing that the derivative of the sum is the sum of the derivatives.)

Therefore

where C is a constant. On the other hand, we have both cot(w) = −cot(−w) and χ(w) = −χ(−w). That is, they are both anti-symmetric. But this can only be true if the constant C = 0. But this implies that our function g satisfies the equation g '(z) = 0; that is, g is a constant. Since

• Complex Analysis: #24 Infinite Products of Functions

For example, we have seen in exercise 6.3 that for z not a real integer, we have the series

converging. In fact it is absolutely convergent. Therefore, according to theorem 47, we must have

converging, and so defining a function, for all z ∉ ℤ. But is this function meromorphic? It is a small exercise to see that the sequence of partial products is uniformly convergent on compact subsets of  ℂ \ ℤ, and thus according to Weierstrass’ convergence theorem, the function is analytic in ℂ \ ℤ. So this example shows one way to proceed in this special case.

On the other hand, polynomials seem to give us a more natural basis for generating analytic functions. Given a polynomial P(z), we can write it as

P(z) = a(a₁ − z) · · · (a_n − z),

where a₁, . . . , a_n are the zeros of the polynomial (perhaps some with multiplicity greater than one), and a ≠ 0 is some constant. If we generalize this to an infinite product, and if we hope that things will converge, then we expect the terms to converge to 1. So it is natural to write something like this:

If the sequence of absolute values |a_n| grows sufficiently rapidly, then we might expect to have convergence. But even a simple sequence like a_n = n does not satisfy this property. Such considerations led to Weierstrass’ product theorem.


Theorem 48 (Weierstrass)
Let (a_n)_n∈ℕ be a sequence of complex numbers with a_n ≠ 0 for all n, and lim_n→∞ a_n = ∞. Then there exist integers m_n  ≥ 0 such that the product

In fact, we can be more specific than this.


Corollary
Let f : ℂ → ℂ be a non-constant entire function. Then there exists an entire function g such that

we have that the function given by φ(z) = f(z)/Ψ(z) is an entire function (with removable singularities) with φ(z) ≠ 0, for all z ∈ ℂ. But then φ'(z)/φ(z) is also an entire function, with antiderivative g(z), say. Let’s look at the derivative of the function φ(z)e^−g(z). We get

So φ(z)e^−g(z) is a constant, which we can absorb into g in such a way that φ(z)e^−g(z) = 1

A simple consequence of this theorem is that every meromorphic function defined throughout ℂ is a rational function. That is, let f be meromorphic, with poles a₁, a₂, . . . . [ Of course if there are only finitely many poles, then this is trivial. On the other hand, since the poles are isolated, there can be at most countably many of them.] Regardless of the way the poles are numbered, since they are isolated we must have lim_n→∞ a_n = ∞. Therefore take the product Ψ · f, where Ψ is an entire function with the same set of zeros as f has poles (weighted according to their orders). Thus Ψ · f is — apart from a set of isolated, removable singularities — an entire function; call it g : ℂ → ℂ. Therefore we can write f = g/Ψ.

Looking at the the formula in Weierstrass’ theorem, we see that it can become a bit of a mess, particularly then the numbers m_n get larger and larger. Things are nicer if there exists a single number m, such that Weierstrass’ theorem works with all m_n ≤ m.

Definition 16
Given a countable set of non-zero complex numbers {a_n}, if there exists an integer m ≥ 0 such that the product

converges, then this is called the canonical product associated with the sequence (a_n)_n∈ℕ. The smallest such m is called the genus of the canonical product.

When does the canonical product exist? That is to say, for a fixed non-negative integer m, and for each a_n, we have a “remainder term” of the form

• Complex Analysis: #23 Infinite Products

After thinking about Weierstrass’ theorem, where we are interested in infinite sums of analytic functions, the question comes up, is it also possible to deal with infinite products? Well it certainly is possible, and this is the subject of a number of classical theorems within complex analysis.

Before we get involved with infinite products of functions, we should first think about something easier, namely infinite products of numbers alone. So let z₁, z₂ , . . . be a sequence of numbers. These give rise to a sequence of “partial products”

But we should realize that there are some special things to think about here which make things different from the simpler situation with partial sums.

• For example with sums, the convergence of the series is not affected if we change a single term. But with products, if one of the terms z_k is changed to 0, then obviously all of the subsequent P_n are zero, regardless of what the further terms look like. Therefore we see that it only makes sense to consider products where all terms are non-zero.


• Another thing is that we could have lim_n→∞ P_n = 0. While this may not seem to be particularly objectionable at first, it is when one realizes that in this case, the limit again remains unchanged if various terms in the product are changed.

Both of these considerations show that, in a way, the number 0 in a product creates the same problems as does the number ∞ in a sum. So we will just agree to do away with the number zero when thinking about infinite products. However, because some people still find it nice to think about the number zero, the following definition will be used.

Definition 15
Let (z_n)_n∈ℕ be a sequence of complex numbers which contains at most finitely many zeros. If the sequence of partial products of the non-zero terms converges to a number which is not zero, then we will say that the infinite product is convergent.

It is a rather trivial observation that, for a convergent product of the form ∏_n∈ℕ z_n, we must have lim_n→∞ z_n = 1. Furthermore, we can assume that at most one of the z_n is a negative real number. For it is obvious that if we have two negative numbers, then it is simpler to just take the corresponding positive numbers. In fact, for this reason it is best to simply exclude negative real numbers from our considerations here completely, and if, as a very special case, we find it convenient to multiply things with the number −1, then that can be done at the end of our calculations.

This means that if we multiply numbers of the form z_n, then we will assume that we can write z_n = r_ne^iθ_n, with −π < θ_n < π. Or put another way, we can write log z_n = log r + iθ_n. This is the principal branch of the logarithm.


Theorem 46
Let z_k = x_k + iy_k for all k ∈ ℕ such that if y_k = 0 then x_k > 0. (That is, all complex numbers are allowed except for real numbers which are not positive.) Then we have that ∏ z_k , (with k = 1, . . ., ∞) is convergent if and only if ∑log z_k , (with k = 1, . . ., ∞) is convergent (where, of course, we take the principal branch of the logarithm).

Proof
First assume that the sum of the logarithms converge. For example, let

Of course, the logarithm always seems troublesome. Therefore the following theorem reduces things to a level which can be more easily checked.


Theorem 47
Writing z_k = 1+a_k, we have ∏ (1+a_k) , (with k = 1, . . ., ∞) is absolutely convergent (that is, the sum of the absolute values of the logarithms, ∑ |log (1+a_k)| , (with k = 1, . . ., ∞) is convergent) if and only if ∑ |a_k| , (with k = 1, . . ., ∞) converges.

Proof
Begin by observing that since we have log '(z) = 1/z, it follows that

• Complex Analysis: #22 Montel`s Theorem

Thinking about Weierstrass’ convergence theorem (theorem 28), let us again consider sequences of functions.

Definition 14
Let G ⊂ ℂ be a region, and for each n ∈ ℕ let f_n : G → ℂ be analytic. This gives us a sequence of functions on G. We will say the sequence is locally bounded if for all z ∈ G, there exists an (open) neighborhood z ∈ U ⊂ G and an M > 0, such that |f_n(w)| ≤ M, for all w ∈ U and all n ∈ ℕ.


Theorem 43
Let f_n : G → ℂ be a locally bounded sequence of analytic functions. Assume there exists a dense subset T ⊂ G, such that (f_n(z))_n∈ℕ is a convergent sequence in ℂ for all z ∈ T. Then there exists an analytic function f : G → ℂ such that f_n → f uniformly on every compact subset of G.

Proof
Begin by observing that we only need prove that for every z₀ ∈ G, there exists an r > 0 such that the sequence of functions f_n is uniformly convergent on B(z₀ , r) (the open disc of radius r centered on z₀). This follows, since given any compact subset K ⊂ G, it can be covered with finitely many such discs.

So given some z₀ ∈ G, we would like to show that there exists an r > 0 such that for all ∈ > 0 there exists an N₀ ∈ ℕ such that |f_n(z) − f_m(z)| < for all n, m ≥ N₀ and |z − z₀| < r. Given this, then for each such z we would have (f_n(z))_n∈ℕ being a Cauchy sequence, converging to a point f(z) in ℂ. Thus the sequence would be uniformly convergent in B(z₀, r) to the function f.

In order to find such an r and N₀, let us begin by using the property that the sequence of functions is locally bounded. Thus there is some M > 0 and an r > 0 such that |f_n(z)| ≤ M for all z ∈ D(z₀, 2r) = {z ∈ G : |z − z₀| ≤ 2r}. Since T is dense in G, and D(z₀, r) is compact, we can find some finite number of points of T in B(z₀, r), call them a₁, . . . , a_k ∈ T, with

Note that the inequality in the third line follows because the radius of the circle is 2r, and of course r is greater than both |ζ − z| and |ζ − a₁|. Also the last inequality is due to the fact that we have assumed that |z − a₁| < ∈ r/6M.


Theorem 44 (Montel)
Assume f_n : G → ℂ (with n ∈ ℕ) is a locally bounded sequence of analytic functions. Then there exists a sub-sequence which is uniformly convergent on every compact subset of G.

Proof
Take some arbitrary sequence {a₁, a₂, . . . } which is dense in G. Since the sequence of points (f_n(a₁))_n∈ℕ is bounded, there exists a convergent sub-sequence, giving a sub-sequence (f_1n)_n∈ℕ of the sequence of functions. Next look at the sequence of points (f_1n(a₁))_n∈ℕ. Again, there is a convergent sub-sequence. And so forth. So for each m ∈ ℕ, we obtain a sequence of functions (f_mn)_n∈ℕ which is such that the sequence of points (f_mn(a₁))_n∈ℕ converges, for all l ≤ m. Therefore the sequence of functions (f_nn)_n∈ℕ satisfies the conditions of theorem 43.

We can use this theorem to find a criterion for the convergence of a sequence of functions as follows.

 
Theorem 45
Again, let f_n : G → ℂ be a locally bounded sequence of analytic functions. Assume there exists some z₀ ∈ G such that the sequences (f_n^(k)(z₀))_n∈ℕ (that is, the sequences of k-th derivatives) converge, for all k. Then (f_n)_n∈ℕ is uniformly convergent on all compact subsets of G.

Proof
According to theorem 43, if (f_n(z))_n∈ℕ is convergent for all z ∈ G, then the sequence of functions is uniformly convergent on all compact subsets of G, and we are finished. So let’s assume that there exists some a ∈ G, such that the sequence of points (f_n(a))_n∈ℕ is not convergent. But at least it must be bounded, so there must be two different sub-sequences of the sequence of functions, being convergent at a to two different values, say one sub-sequence converges to the value v_a ∈ ℕ and the other converges to w_a, where w_a ≠ v_a. However, using Montel’s theorem, we have sub-sequences of these sub-sequences of functions, converging to two different analytic functions: f, g : G → ℂ, with f(a) = w_a ≠ v_a = g(a). Looking at the point z₀, we have

lim_n→∞  f_n^(k)(z₀) = f^(k)(z₀) = g^(k)(z₀),

that is, (f − g)^k(z₀) = 0 for all k. Therefore, the function f − g is zero in a neighborhood of z₀, but this implies that it is zero everywhere, including at the point a, which gives us a contradiction.